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Functions Representations
Lecture 9, CMSC 56
Allyn Joy D. Calcaben
Functions
Definition 1
Let A and B be two sets. A function f from A to B, denoted f : A
→ B , is an assignment of exactly one element of B to each element
of A. We write f(a) = b if b is the unique element of B assigned by
the function f to the element a of A.
Functions
Definition 1
Let A and B be two sets. A function f from A to B, denoted f : A
→ B , is an assignment of exactly one element of B to each element
of A. We write f(a) = b if b is the unique element of B assigned by
the function f to the element a of A.
Functions are sometimes also called mappings or transformations
Functions
A f: A → B B
Functions
A f: A → B B
Functions
X
NOT ALLOWED !!
Definition 2
If f is a function from A to B, we say that A is the domain of f
and B is the codomain of f. If f(a) = b, we say that b is the image of
a and a is the preimage of b. The range, or image, of f is the set of
all images of elements of A. Also, if f is a function from A to B, we
say that f maps A to B.
Functions
What are the domain, codomain, and range of the function that
assigns grades to students in a discrete mathematics class. Suppose
that the grades are 1.00 for Ronn, 3.00 for Ricca, 1.75 for Robien,
2.50 for Zin, 3.00 for Angelique, and 1.50 for Mace.
Example
If f is a function specified by R, then
f(x) is the grade of x, where x is a student.
Solution
If f is a function specified by R, then
f(x) is the grade of x, where x is a student.
f(Ronn) = 1.00
f(Ricca) = 3.00
f(Robien) = 1.75
f(Zin) = 2.50
f(Mace) = 1.50
f(Angelique) = 3.00
Solution
If f is a function specified by R, then
f(x) is the grade of x, where x is a student.
f(Ronn) = 1.00 Ronn •
f(Ricca) = 3.00 Ricca •
f(Robien) = 1.75 Robien •
f(Zin) = 2.50 Zin •
f(Mace) = 1.50 Mace •
f(Angelique) = 3.00 Angelique •
Solution
• 1.00
• 1.25
• 1.50
• 1.75
• 2.00
• 2.25
• 2.50
• 2.75
• 3.00
Set A (Domain)
{Ronn, Ricca, Robien, Zin, Angelique, Mace}
Set B (Codomain)
{1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2.50, 2.75, 3.0}
Range
{1.00, 1.50, 1.75, 2.50, 3.0}
Solution
Let R be the relation with ordered pairs (Chloei, 21), (Ezekiel, 22),
(Hannah, 24), (Jasper, 21), (Jayven, 23), (Kent, 22), (Maren, 23),
(Robyn, 24), (Sean, 22), and (Vinze, 25). Here each pair consists of a
graduate student and their age. Specify a function determined by
this relation.
Challenge (1/4 Sheet Paper)
If f is a function specified by R, then
f(x) is the age of x, where x is a student.
Solution
If f is a function specified by R, then
f(x) is the age of x, where x is a student.
f(Chloei) = 21 f(Ezekiel) = 22 f(Hannah) = 24
f(Jasper) = 21 f(Jayven) = 23 f(Kent) = 22
f(Maren) = 23 f(Robyn) = 24 f(Sean) = 22
f(Vinze) = 25.
Solution
If f is a function specified by R, then
f(x) is the age of x, where x is a student.
Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren,
Robyn, Sean, Vinze }
Solution
If f is a function specified by R, then
f(x) is the age of x, where x is a student.
Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren,
Robyn, Sean, Vinze }
Set B (Codomain) = { Z+ | f(x) < 100 }
Solution
If f is a function specified by R, then
f(x) is the age of x, where x is a student.
Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren,
Robyn, Sean, Vinze }
Set B (Codomain) = { Z+ | f(x) < 100 }
Range = { 21, 22, 23, 24, 25 }
Solution
Definition 3
If f1 and f2 be functions from A to R. Then f1 + f2 and f1 f2 are
also functions from A to R defined for all x ϵ A by
( f1 + f2 )(x) = f1(x) + f2 (x)
( f1 f2 )(x) = f1(x) f2 (x)
Functions
Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) =
x – x2. What are the functions f1 + f2 and f1 f2 ?
Example
Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) =
x – x2. What are the functions f1 + f2 and f1 f2 ?
( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x
Solution
Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) =
x – x2. What are the functions f1 + f2 and f1 f2 ?
( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x
( f1 f2 )(x) = f1(x) f2 (x) = x2 (x – x2) = x3 – x4
Solution
Definition 4
Let f be a function from A to B and let S be a subset of A. The
image of S under the function f is the subset of B that consists of
the images of the elements of S. We denote the image of S by f(S),
so
f(S) = { t | Ǝs ϵ S ( t = f(s) ) }
Functions
Definition 4
Let f be a function from A to B and let S be a subset of A. The
image of S under the function f is the subset of B that consists of
the images of the elements of S. We denote the image of S by f(S),
so
f(S) = { t | Ǝs ϵ S ( t = f(s) ) }
We also use the shorthand { f(s) | s ϵ S } to denote this set.
Functions
One-to-One
and Onto functions
Definition 5
A function f is said to be one-to-one, or an injunction, if and
only if f(a) = f(b) implies that a = b for all a and b in the domain of f.
A function is said to be injective if it is one-to-one.
Injuction Functions
Definition 5
A function f is said to be one-to-one, or an injunction, if and
only if f(a) = f(b) implies that a = b for all a and b in the domain of f.
A function is said to be injective if it is one-to-one.
We can express that f is one-to-one using quantifiers as ꓯaꓯb( f(a)
= f(b) → a = b ) or equivalent ꓯaꓯb( a ≠ b → f(a) ≠ f(b) ), where
the universe of discourse is the domain of the function.
Injuction Functions
Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 }
with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one.
Example
Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 }
with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one.
a • • 1
b • • 2
c • • 3
d • • 4
• 5
Solution
Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 }
with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one.
a • • 1
b • • 2
c • • 3
d • • 4
• 5
Solution
The function f is one-to-one
because f takes on different
values at the four elements
of its domain.
Definition 6
A function f whose domain and codomain are subsets of the
set of real numbers is called increasing if f(x) ≤ f(y), and strictly
increasing if f(x) < f(y), whenever x < y and x and y are in the
domain of f.
Increasing & Decreasing Functions
Definition 6
Similarly, f is called decreasing if f(x) ≥ f(y), and strictly
decreasing if f(x) > f(y), whenever x < y and x and y are in the
domain of f.
Increasing & Decreasing Functions
Definition 6
Similarly, f is called decreasing if f(x) ≥ f(y), and strictly
decreasing if f(x) > f(y), whenever x < y and x and y are in the
domain of f.
NOTE:
Strictly increasing and strictly decreasing functions are one-to-one.
Increasing & Decreasing Functions
Definition 7
A function f from A to B is called onto, or a surjection, if and
only if for every element b ϵ B there is an element a ϵ A with f(a) =
b. A function f is called surjective if it is onto.
Surjection Functions
Definition 7
A function f from A to B is called onto, or a surjection, if and
only if for every element b ϵ B there is an element a ϵ A with f(a) =
b. A function f is called surjective if it is onto.
Remark: A function f is onto if ∀y∃x( f(x) = y ), where the domain
for x is the domain of the function and the domain for y is the
codomain of the function.
Surjection Functions
Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by
f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function?
Example
Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by
f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function?
a • • 1
b • • 2
c • • 3
d •
Solution
Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by
f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function?
a • • 1
b • • 2
c • • 3
d •
Solution
Since all three elements of
the codomain are images of
elements in the domain, we
see that f is onto.
Definition 8
The function f is a one-to-one correspondence, or a bijection,
if it is both one-to-one and onto. We also say that such a function is
bijective.
Bijective Functions
Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4,
f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection?
Example
Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4,
f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection?
a • • 1
b • • 2
c • • 3
d • • 4
Example
Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4,
f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection?
a • • 1
b • • 2
c • • 3
d • • 4
Example
Since the function f is one-to-one
and onto, then function f is a
bisection.
Let A be a set. The identity function on A is the function ιA : A → A,
where
ιA(x) = x
for all x ϵ A. In other words, the identity function ιA is the function
that assigns each element to itself. The function ιA is one-to-one
and onto, so it is a bijection.
Identity Function
Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)?
Example
Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)?
ιA(1) = 1
ιA(2) = 2
ιA(3) = 3
Solution
Inverse Functions
Definition 9
Let f be a one-to-one correspondence from the set A to the
set B. The inverse function of f is the function that assigns to an
element b belonging to B the unique element a in A such that f(a) =
b. The inverse function of f is denoted by f -1. Hence, f -1(b) = a
when f(a) = b.
Inverse Functions
Be sure not to confuse the function f -1 with the function 1 / f , which is
the function that assigns to each x in the domain the value 1 / f(x). Notice
that the latter makes sense only when f(x) is a non-zero real number.
Inverse Functions
Be sure not to confuse the function f -1 with the function 1 / f , which is
the function that assigns to each x in the domain the value 1 / f(x). Notice
that the latter makes sense only when f(x) is a non-zero real number.
A one-to-one correspondence is called invertible because we can
define an inverse of this function. A function is not invertible if it is not a one-
to-one correspondence, because the inverse of such a function does not
exist.
Inverse Functions
Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3,
and f(c) = 1. Is f invertible, and if it is, what is its inverse?
Example
Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3,
and f(c) = 1. Is f invertible, and if it is, what is its inverse?
a • • 1
b • • 2
c • • 3
Solution
Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3,
and f(c) = 1. Is f invertible, and if it is, what is its inverse?
a • • 1 f -1(1) = c
b • • 2 f -1(2) = a
c • • 3 f -1(3) = b
Solution
Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3,
and f(c) = 1. Is f invertible, and if it is, what is its inverse?
a • • 1 f -1(1) = c
b • • 2 f -1(2) = a
c • • 3 f -1(3) = b
Solution
The function f is invertible because it is a one-to-one correspondence.
Let f be the function from R to R with f(x) = x2. Is f invertible?
Example
Let f be the function from R to R with f(x) = x2. Is f invertible?
Because f(-2) = f(2) = 4, f is not one-to-one. If an inverse function were
defined, it would have to assign two elements to 4. Hence, f is not invertible.
Solution
Composition of Functions
Definition 10
Let g be a function from the set A to the set B and let f be a
function from the set B to the set C. The composition of the
functions f and g, denoted for all a ϵ A by f ◦ g, is defined by
(f ◦ g)(a) = f(g(a))
Composition of Functions
Let g be the function from the set { a, b, c } to itself such that g(a) =
b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c }
to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is
the composition of f and g, and what is the composition of g and f?
Example
Let g be the function from the set { a, b, c } to itself such that g(a) =
b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c }
to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is
the composition of f and g, and what is the composition of g and f?
Composition of f and g:
(f ◦ g)(a) = f(g(a)) = f(b) = 2
(f ◦ g)(b) = f(g(b)) = f(c) = 1
(f ◦ g)(c) = f(g(c)) = f(a) = 3
Solution
Let g be the function from the set { a, b, c } to itself such that g(a) =
b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c }
to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is
the composition of f and g, and what is the composition of g and f?
Composition of f and g: Composition of g and f:
(f ◦ g)(a) = f(g(a)) = f(b) = 2
(f ◦ g)(b) = f(g(b)) = f(c) = 1
(f ◦ g)(c) = f(g(c)) = f(a) = 3
Solution
g ◦ f is not defined, because the
range of f is not a subset of the
domain g.
Let f and g be the functions from the set of integers to the set of integers
defined by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g?
What is the composition of g and f?
Challenge (1/4 Sheet Paper)
Let f and g be the functions from the set of integers to the set of integers
defined by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g?
What is the composition of g and f?
Composition of f and g:
(f ◦ g)(x) = f(g(x)) = f(3x + 2) = 2(3x + 2) + 3 = 6x + 7
Composition of g and f:
(g ◦ f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 2 = 6x + 11
Solution
Any Questions?
EXAM RESULT
46
37
42
2
35
7
A B C D E F
88.85%
70.52%
82.92%
28.13%
71.88%
36.11%
A B C D E F
Passed,…
Failed,
17.02%
Best Student Performance with only
11.15% FAILURE
Lowest Student Performance with almost
72% FAILURE
CMSC 56
1st Long Exam| AY 2018 – 2019
EXAM PARTSHistogram
Students passed in Part
A
08Students failed
39Students passed
Students failed in Part
D
46
45
PARTA
PARTD
ANSWER KEYON SELECTED PARTS
PARTD
1. Thereissomeoneinthis classwhodoesnothavea
goodattitude.
x = “studentinthisclass”
A(x)= “x hasagoodattitude”
a. Ǝx ~A(x)OriginalStatement
b. ~(Ǝx~A(x))= ꓯxA(x)
“Everyoneinthisclasshasa goodattitude.”
2. Norabbitknows Calculus.
x = “rabbit”
C(x)= “x knowscalculus”
a.~ƎxC(x)OriginalStatement
b. ~(~ƎxC(x)) = Ǝx A(x)
“Thereisa rabbitthat knows Calculus.”
PARTF
1. I driveifit istoo dangerousortoo fartobike.
p = “Idrive”
q = “Itistoo dangerous”
r = “Itistoo fartobike”
(qV r)→ p OriginalStatement
a. p → (qV r)Converse
“IfI drive,thenit istoo dangerousor too
fartobike.”
“Itistoo dangerousortoo fartobike ifI
drive.”
b. ~p→~(qV r)Contrapositive
“IfI don’t drive,thenitisnottoo
dangerousortoo fartobike.”
“IfI don’t drive,thenitisnottoo
dangerousandnottoofartobike.”
“Itisnottoodangerousortoo farto bike
ifIdon’t drive.”
“Itisnottoodangerousand nottoo farto
bike if I don’tdrive.”
2.
a. Nostudentinyourclasshastaken acourseinlogic
programming.
x = “studentsinthisclass”
L(x)= “x hastakena coursein
logicprogramming”
ꓯx~L(x)
b. ComputerScienceis easyorcampingisfun, aslong
asitissunnyand thehomework isdone.
m =“computerscienceiseasy”
c = “campingisfun”
s = “itissunny”
h = “homeworkisdone”
(sꓥ h) ↔ ( m V c )
(sꓥ h) → ( m V c )
100.0% Garcia, Patricia Marie
96.0% Samson, Aron Miles
93.0% Honeyman, John
91.0% Jomoc, Gracielou
Patot, Oliver
90.0% Garcia, Michael John
CMSC 56 1st LE Top 5 Scorers

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CMSC 56 | Lecture 9: Functions Representations

  • 1. Functions Representations Lecture 9, CMSC 56 Allyn Joy D. Calcaben
  • 3. Definition 1 Let A and B be two sets. A function f from A to B, denoted f : A → B , is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. Functions
  • 4. Definition 1 Let A and B be two sets. A function f from A to B, denoted f : A → B , is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. Functions are sometimes also called mappings or transformations Functions
  • 5. A f: A → B B Functions
  • 6. A f: A → B B Functions X NOT ALLOWED !!
  • 7. Definition 2 If f is a function from A to B, we say that A is the domain of f and B is the codomain of f. If f(a) = b, we say that b is the image of a and a is the preimage of b. The range, or image, of f is the set of all images of elements of A. Also, if f is a function from A to B, we say that f maps A to B. Functions
  • 8. What are the domain, codomain, and range of the function that assigns grades to students in a discrete mathematics class. Suppose that the grades are 1.00 for Ronn, 3.00 for Ricca, 1.75 for Robien, 2.50 for Zin, 3.00 for Angelique, and 1.50 for Mace. Example
  • 9. If f is a function specified by R, then f(x) is the grade of x, where x is a student. Solution
  • 10. If f is a function specified by R, then f(x) is the grade of x, where x is a student. f(Ronn) = 1.00 f(Ricca) = 3.00 f(Robien) = 1.75 f(Zin) = 2.50 f(Mace) = 1.50 f(Angelique) = 3.00 Solution
  • 11. If f is a function specified by R, then f(x) is the grade of x, where x is a student. f(Ronn) = 1.00 Ronn • f(Ricca) = 3.00 Ricca • f(Robien) = 1.75 Robien • f(Zin) = 2.50 Zin • f(Mace) = 1.50 Mace • f(Angelique) = 3.00 Angelique • Solution • 1.00 • 1.25 • 1.50 • 1.75 • 2.00 • 2.25 • 2.50 • 2.75 • 3.00
  • 12. Set A (Domain) {Ronn, Ricca, Robien, Zin, Angelique, Mace} Set B (Codomain) {1.00, 1.25, 1.50, 1.75, 2.00, 2.25, 2.50, 2.75, 3.0} Range {1.00, 1.50, 1.75, 2.50, 3.0} Solution
  • 13. Let R be the relation with ordered pairs (Chloei, 21), (Ezekiel, 22), (Hannah, 24), (Jasper, 21), (Jayven, 23), (Kent, 22), (Maren, 23), (Robyn, 24), (Sean, 22), and (Vinze, 25). Here each pair consists of a graduate student and their age. Specify a function determined by this relation. Challenge (1/4 Sheet Paper)
  • 14. If f is a function specified by R, then f(x) is the age of x, where x is a student. Solution
  • 15. If f is a function specified by R, then f(x) is the age of x, where x is a student. f(Chloei) = 21 f(Ezekiel) = 22 f(Hannah) = 24 f(Jasper) = 21 f(Jayven) = 23 f(Kent) = 22 f(Maren) = 23 f(Robyn) = 24 f(Sean) = 22 f(Vinze) = 25. Solution
  • 16. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Solution
  • 17. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Set B (Codomain) = { Z+ | f(x) < 100 } Solution
  • 18. If f is a function specified by R, then f(x) is the age of x, where x is a student. Set A (Domain) = { Chloei, Ezekiel, Hannah, Jasper, Jayven, Kent, Maren, Robyn, Sean, Vinze } Set B (Codomain) = { Z+ | f(x) < 100 } Range = { 21, 22, 23, 24, 25 } Solution
  • 19. Definition 3 If f1 and f2 be functions from A to R. Then f1 + f2 and f1 f2 are also functions from A to R defined for all x ϵ A by ( f1 + f2 )(x) = f1(x) + f2 (x) ( f1 f2 )(x) = f1(x) f2 (x) Functions
  • 20. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? Example
  • 21. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? ( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x Solution
  • 22. Let f1 and f2 be functions from R to R such that f1(x) = x2 and f2(x) = x – x2. What are the functions f1 + f2 and f1 f2 ? ( f1 + f2 )(x) = f1(x) + f2 (x) = x2 + (x – x2) = x ( f1 f2 )(x) = f1(x) f2 (x) = x2 (x – x2) = x3 – x4 Solution
  • 23. Definition 4 Let f be a function from A to B and let S be a subset of A. The image of S under the function f is the subset of B that consists of the images of the elements of S. We denote the image of S by f(S), so f(S) = { t | Ǝs ϵ S ( t = f(s) ) } Functions
  • 24. Definition 4 Let f be a function from A to B and let S be a subset of A. The image of S under the function f is the subset of B that consists of the images of the elements of S. We denote the image of S by f(S), so f(S) = { t | Ǝs ϵ S ( t = f(s) ) } We also use the shorthand { f(s) | s ϵ S } to denote this set. Functions
  • 26. Definition 5 A function f is said to be one-to-one, or an injunction, if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. A function is said to be injective if it is one-to-one. Injuction Functions
  • 27. Definition 5 A function f is said to be one-to-one, or an injunction, if and only if f(a) = f(b) implies that a = b for all a and b in the domain of f. A function is said to be injective if it is one-to-one. We can express that f is one-to-one using quantifiers as ꓯaꓯb( f(a) = f(b) → a = b ) or equivalent ꓯaꓯb( a ≠ b → f(a) ≠ f(b) ), where the universe of discourse is the domain of the function. Injuction Functions
  • 28. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. Example
  • 29. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. a • • 1 b • • 2 c • • 3 d • • 4 • 5 Solution
  • 30. Determine whether the function f from { a, b, c, d } to { 1, 2, 3, 4, 5 } with f(a) = 4, f(b) = 5, f(c) = 1, and f(d) = 3 is one-to-one. a • • 1 b • • 2 c • • 3 d • • 4 • 5 Solution The function f is one-to-one because f takes on different values at the four elements of its domain.
  • 31. Definition 6 A function f whose domain and codomain are subsets of the set of real numbers is called increasing if f(x) ≤ f(y), and strictly increasing if f(x) < f(y), whenever x < y and x and y are in the domain of f. Increasing & Decreasing Functions
  • 32. Definition 6 Similarly, f is called decreasing if f(x) ≥ f(y), and strictly decreasing if f(x) > f(y), whenever x < y and x and y are in the domain of f. Increasing & Decreasing Functions
  • 33. Definition 6 Similarly, f is called decreasing if f(x) ≥ f(y), and strictly decreasing if f(x) > f(y), whenever x < y and x and y are in the domain of f. NOTE: Strictly increasing and strictly decreasing functions are one-to-one. Increasing & Decreasing Functions
  • 34. Definition 7 A function f from A to B is called onto, or a surjection, if and only if for every element b ϵ B there is an element a ϵ A with f(a) = b. A function f is called surjective if it is onto. Surjection Functions
  • 35. Definition 7 A function f from A to B is called onto, or a surjection, if and only if for every element b ϵ B there is an element a ϵ A with f(a) = b. A function f is called surjective if it is onto. Remark: A function f is onto if ∀y∃x( f(x) = y ), where the domain for x is the domain of the function and the domain for y is the codomain of the function. Surjection Functions
  • 36. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? Example
  • 37. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? a • • 1 b • • 2 c • • 3 d • Solution
  • 38. Let f be the function from { a, b, c, d } to { 1, 2, 3 } defined by f(a) = 3, f(b) = 2, f(c) = 1, and f(d) = 3. Is f an onto function? a • • 1 b • • 2 c • • 3 d • Solution Since all three elements of the codomain are images of elements in the domain, we see that f is onto.
  • 39. Definition 8 The function f is a one-to-one correspondence, or a bijection, if it is both one-to-one and onto. We also say that such a function is bijective. Bijective Functions
  • 40. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? Example
  • 41. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? a • • 1 b • • 2 c • • 3 d • • 4 Example
  • 42. Let f be the function from { a, b, c, d } to { 1, 2, 3, 4 } with f(a) = 4, f(b) = 2, f(c) = 1, and f(d) = 3. Is f a bijection? a • • 1 b • • 2 c • • 3 d • • 4 Example Since the function f is one-to-one and onto, then function f is a bisection.
  • 43.
  • 44. Let A be a set. The identity function on A is the function ιA : A → A, where ιA(x) = x for all x ϵ A. In other words, the identity function ιA is the function that assigns each element to itself. The function ιA is one-to-one and onto, so it is a bijection. Identity Function
  • 45. Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)? Example
  • 46. Let A = { 1, 2, 3 }, what is ιA(1)? ιA(2)? ιA(3)? ιA(1) = 1 ιA(2) = 2 ιA(3) = 3 Solution
  • 47.
  • 49. Definition 9 Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f is the function that assigns to an element b belonging to B the unique element a in A such that f(a) = b. The inverse function of f is denoted by f -1. Hence, f -1(b) = a when f(a) = b. Inverse Functions
  • 50. Be sure not to confuse the function f -1 with the function 1 / f , which is the function that assigns to each x in the domain the value 1 / f(x). Notice that the latter makes sense only when f(x) is a non-zero real number. Inverse Functions
  • 51. Be sure not to confuse the function f -1 with the function 1 / f , which is the function that assigns to each x in the domain the value 1 / f(x). Notice that the latter makes sense only when f(x) is a non-zero real number. A one-to-one correspondence is called invertible because we can define an inverse of this function. A function is not invertible if it is not a one- to-one correspondence, because the inverse of such a function does not exist. Inverse Functions
  • 52. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? Example
  • 53. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 b • • 2 c • • 3 Solution
  • 54. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 f -1(1) = c b • • 2 f -1(2) = a c • • 3 f -1(3) = b Solution
  • 55. Let f be the function from { a, b, c } to { 1, 2, 3 } such that f(a) = 2, f(b) = 3, and f(c) = 1. Is f invertible, and if it is, what is its inverse? a • • 1 f -1(1) = c b • • 2 f -1(2) = a c • • 3 f -1(3) = b Solution The function f is invertible because it is a one-to-one correspondence.
  • 56. Let f be the function from R to R with f(x) = x2. Is f invertible? Example
  • 57. Let f be the function from R to R with f(x) = x2. Is f invertible? Because f(-2) = f(2) = 4, f is not one-to-one. If an inverse function were defined, it would have to assign two elements to 4. Hence, f is not invertible. Solution
  • 59. Definition 10 Let g be a function from the set A to the set B and let f be a function from the set B to the set C. The composition of the functions f and g, denoted for all a ϵ A by f ◦ g, is defined by (f ◦ g)(a) = f(g(a)) Composition of Functions
  • 60.
  • 61. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Example
  • 62. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Composition of f and g: (f ◦ g)(a) = f(g(a)) = f(b) = 2 (f ◦ g)(b) = f(g(b)) = f(c) = 1 (f ◦ g)(c) = f(g(c)) = f(a) = 3 Solution
  • 63. Let g be the function from the set { a, b, c } to itself such that g(a) = b, g(b) = c, and g(c) = a. Let f be the function from the set { a, b, c } to the set { 1, 2, 3 } such that f(a) = 3, f(b) = 2, and f(c) = 1. What is the composition of f and g, and what is the composition of g and f? Composition of f and g: Composition of g and f: (f ◦ g)(a) = f(g(a)) = f(b) = 2 (f ◦ g)(b) = f(g(b)) = f(c) = 1 (f ◦ g)(c) = f(g(c)) = f(a) = 3 Solution g ◦ f is not defined, because the range of f is not a subset of the domain g.
  • 64. Let f and g be the functions from the set of integers to the set of integers defined by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the composition of g and f? Challenge (1/4 Sheet Paper)
  • 65. Let f and g be the functions from the set of integers to the set of integers defined by f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the composition of g and f? Composition of f and g: (f ◦ g)(x) = f(g(x)) = f(3x + 2) = 2(3x + 2) + 3 = 6x + 7 Composition of g and f: (g ◦ f)(x) = g(f(x)) = g(2x + 3) = 3(2x + 3) + 2 = 6x + 11 Solution
  • 68. 46 37 42 2 35 7 A B C D E F 88.85% 70.52% 82.92% 28.13% 71.88% 36.11% A B C D E F Passed,… Failed, 17.02% Best Student Performance with only 11.15% FAILURE Lowest Student Performance with almost 72% FAILURE CMSC 56 1st Long Exam| AY 2018 – 2019 EXAM PARTSHistogram Students passed in Part A 08Students failed 39Students passed Students failed in Part D 46 45 PARTA PARTD
  • 69. ANSWER KEYON SELECTED PARTS PARTD 1. Thereissomeoneinthis classwhodoesnothavea goodattitude. x = “studentinthisclass” A(x)= “x hasagoodattitude” a. Ǝx ~A(x)OriginalStatement b. ~(Ǝx~A(x))= ꓯxA(x) “Everyoneinthisclasshasa goodattitude.” 2. Norabbitknows Calculus. x = “rabbit” C(x)= “x knowscalculus” a.~ƎxC(x)OriginalStatement b. ~(~ƎxC(x)) = Ǝx A(x) “Thereisa rabbitthat knows Calculus.” PARTF 1. I driveifit istoo dangerousortoo fartobike. p = “Idrive” q = “Itistoo dangerous” r = “Itistoo fartobike” (qV r)→ p OriginalStatement a. p → (qV r)Converse “IfI drive,thenit istoo dangerousor too fartobike.” “Itistoo dangerousortoo fartobike ifI drive.” b. ~p→~(qV r)Contrapositive “IfI don’t drive,thenitisnottoo dangerousortoo fartobike.” “IfI don’t drive,thenitisnottoo dangerousandnottoofartobike.” “Itisnottoodangerousortoo farto bike ifIdon’t drive.” “Itisnottoodangerousand nottoo farto bike if I don’tdrive.” 2. a. Nostudentinyourclasshastaken acourseinlogic programming. x = “studentsinthisclass” L(x)= “x hastakena coursein logicprogramming” ꓯx~L(x) b. ComputerScienceis easyorcampingisfun, aslong asitissunnyand thehomework isdone. m =“computerscienceiseasy” c = “campingisfun” s = “itissunny” h = “homeworkisdone” (sꓥ h) ↔ ( m V c ) (sꓥ h) → ( m V c )
  • 70. 100.0% Garcia, Patricia Marie 96.0% Samson, Aron Miles 93.0% Honeyman, John 91.0% Jomoc, Gracielou Patot, Oliver 90.0% Garcia, Michael John CMSC 56 1st LE Top 5 Scorers

Editor's Notes

  1. Note that the functions f1 + f2 and f1f2 have been defined by specifying their values at x in terms of the values of f1 and f2 at x.
  2. The notation f(S) for the image of the set S under the function f is potentially ambiguous. Here, f(S)denotes a set, and not the value of the function f for the set S.
  3. Why? One-to-one function: A function is one-to-one if and only if f(x)  f(y), whenever x  y.
  4. ι = Greek letter iota
  5. ι = Greek letter iota
  6. ι = Greek letter iota
  7. Function f is not onto.
  8. In other words, f ◦ g is the function that assigns to the element a of A the element assigned by f to g(a). That is, to find (f ◦ g)(a) we first apply the function g to a to obtain g(a) and then we apply the function f to the result g(a) to obtain (f ◦ g)(a) = f(g(a)). Note that the composition f◦ g cannot be defined unless the range of g is a subset of the domain of f . In Figure 7 the composition of functions is shown.
  9. Remark: Note that even though f ◦ g and g ◦ f are defined for the functions f and g in Example 23, f ◦ g and g ◦ f are not equal. In other words, the commutative law does not hold for the composition of functions.